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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #85 : Solving And Graphing Radicals

Solve: $\sqrt[5]{-3x}-3 = -1$

Possible Answers:

$-\frac{32}{3}$

$-\frac{20}{3}$

$\textup{No solution.}$

$-\frac{2}{3}$

$-\frac{16}{3}$

Correct answer:

$-\frac{32}{3}$

Explanation:

Add three on both sides.

$\sqrt[5]{-3x}-3+3 = -1+3$

$\sqrt[5]{-3x} = 2$

Raise both sides by the power of five to eliminate the radical.

$(\sqrt[5]{-3x} )^5= 2^5$

-3x =32

Divide by negative three on both sides.

$\frac{-3x }{-3}=\frac{32}{-3}$

The answer is: $-\frac{32}{3}$

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Example Question #86 : Solving And Graphing Radicals

Solve the equation: $\sqrt5 \cdot \sqrt{2x} -10= -9$

Possible Answers:

$\textup{No solution.}$

$\frac{1}{20}$

$\frac{1}{5}$

$\frac{1}{2}$

$\frac{1}{10}$

Correct answer:

$\frac{1}{10}$

Explanation:

The radicals can be combined as one radical by multiplying the two inner terms together.

The equation becomes:

$\sqrt{10x} -10= -9$

Add 10 on both sides.

$\sqrt{10x} -10+10= -9+10$

Simplify both sides.

$\sqrt{10x} =1$

Square both sides.

10x=1

Divide by ten on both sides.

$\frac{10x}{10}=\frac{1}{10}$

The answer is: $\frac{1}{10}$

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Example Question #401 : Radicals

Solve the radical: $\sqrt{\sqrt{2x}} = \sqrt{8}$

Possible Answers:

256

128

Correct answer:

Explanation:

Rewrite the radicals so that they are in exponential form.

$\sqrt{x} = x^\frac{1}{2}$

$\sqrt{\sqrt{2x}} =[ (2x)^{\frac{1}{2}}]^{\frac{1}{2}} = (2x)^{\frac{1}{4}}$

The equation becomes:

$(2x)^{\frac{1}{4}} = 8^{\frac{1}{2}}$

Raise both sides by the power of four to eliminate the fractional exponents.

$[(2x)^{\frac{1}{4}}] ^4= [8^{\frac{1}{2}}]^4$

2x=64

Divide by two on both sides.

$\frac{2x}{2}=\frac{64}{2}$

The answer is:

Report an Error

Example Question #82 : Solving Radical Equations

Solve: $4=2^{10x-8}$

Possible Answers:

$-\frac{1}{2}$

$\frac{3}{4}$

Correct answer:

Explanation:

Evaluate by changing the base of the left term.

$2^2=2^{10x-8}$

Now that both bases are common, set the powers equal and solve for x.

2=10x-8

Add 8 on both sides.

2+8=10x-8+8

10=10x

Divide by ten on both sides.

$\frac{10}{10}=\frac{10x}{10}$

The answer is:

Report an Error

Example Question #81 : Solving And Graphing Radicals

Evaluate: $\sqrt[3]{2x-8} = 2$

Possible Answers:

$\textup{No solution.}$

-8,8

Correct answer:

Explanation:

Cube both sides to eliminate the radical.

$(\sqrt[3]{2x-8}) ^3= 2^3$

2x-8 =8

Add eight on both sides.

2x-8+(8) =8+(8)

2x=16

Divide by 2 on both sides.

$\frac{2x}{2}=\frac{16}{2}$

The answer is:

Report an Error

Example Question #81 : Solving Radical Equations

Solve: $2\sqrt[3]{2-5x} = 10$

Possible Answers:

$\frac{13}{5}$

$-\frac{123}{5}$

$\frac{123}{5}$

$-\frac{13}{5}$

$\frac{127}{5}$

Correct answer:

$-\frac{123}{5}$

Explanation:

Solve by first dividing both sides by two.

$\frac{2\sqrt[3]{2-5x} }{2}= \frac{10}{2}$

Simplify both sides.

$\sqrt[3]{2-5x} =5$

Cube both sides.

$(\sqrt[3]{2-5x})^3 =5^3$

2-5x =125

Subtract two from both sides.

2-5x -(2)=125-(2)

-5x=123

Divide by negative five on both sides.

$\frac{-5x}{-5}=\frac{123}{-5}$

The answer is: $-\frac{123}{5}$

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Example Question #91 : Solving And Graphing Radicals

Solve: $\sqrt{3x-3}-3 = 3$

Possible Answers:

Correct answer:

Explanation:

Add three on both sides.

$\sqrt{3x-3}-3+3 = 3+3$

The equation becomes:

$\sqrt{3x-3} =6$

Square both sides to eliminate the radical.

$(\sqrt{3x-3} )^2=6^2$

3x-3 = 36

Add three on both sides.

3x-3 +3= 36+3

3x=39

Divide by three on both sides.

$\frac{3x}{3}=\frac{39}{3}$

The answer is:

Report an Error

Example Question #92 : Solving And Graphing Radicals

Solve the equation: $4\sqrt[3]{-2x-3} = 8$

Possible Answers:

$-\frac{3}{2}$

$-\frac{5}{2}$

$\textup{There is no solution.}$

$-\frac{11}{2}$

$-\frac{67}{2}$

Correct answer:

$-\frac{11}{2}$

Explanation:

Divide by four on both sides.

$\frac{4\sqrt[3]{-2x-3} }{4}= \frac{8}{4}$

$\sqrt[3]{-2x-3} = 2$

Cube both sides to eliminate the radical.

$(\sqrt[3]{-2x-3})^3 = 2^3$

-2x-3=8

Add three on both sides.

-2x-3+3=8+3

-2x=11

Divide by negative two on both sides.

$\frac{-2x}{-2}=\frac{11}{-2}$

The answer is: $-\frac{11}{2}$

Report an Error

Example Question #93 : Solving And Graphing Radicals

Solve: $-\sqrt{x+8}-16 = -20$

Possible Answers:

$\textup{No solution.}$

Correct answer:

Explanation:

Add 16 on both sides.

$-\sqrt{x+8}-16+16= -20+16$

$-\sqrt{x+8} = -4$

Divide both sides by negative one.

$\frac{-\sqrt{x+8} }{-1}=\frac{ -4}{-1}$

$\sqrt{x+8}=4$

Square both sides.

$(\sqrt{x+8}) ^2=4^2$

x+8=16

Subtract eight on both sides.

x+8-8=16-8

x=8

The answer is:

Report an Error

Example Question #94 : Solving Radical Equations

Solve: $\sqrt{-2x-5} = 9$

Possible Answers:

$\textup{No solution.}$

Correct answer:

Explanation:

Square both sides in order to eliminate the radical.

$(\sqrt{-2x-5}) ^2= 9^2$

-2x-5 = 81

Add 5 on both sides.

-2x-5 +5= 81+5

-2x=86

Divide by negative two on both sides.

$\frac{-2x}{-2}=\frac{86}{-2}$

Reduce both fractions.

The answer is:

Report an Error

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #85 : Solving And Graphing Radicals

Example Question #86 : Solving And Graphing Radicals

Example Question #401 : Radicals

Example Question #82 : Solving Radical Equations

Example Question #81 : Solving And Graphing Radicals

Example Question #81 : Solving Radical Equations

Example Question #91 : Solving And Graphing Radicals

Example Question #92 : Solving And Graphing Radicals

Example Question #93 : Solving And Graphing Radicals

Example Question #94 : Solving Radical Equations

All Algebra II Resources

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